Question

A circle is inscribed in a triangle ABC. It

touches sides AB, BC and AC at the points

P, Q and R respectively. If BP = 6.5 cm,

CQ = 4.5 Cm and AR = 5.5 cm, then the

perimeter (in cm) of the triangle ∆ABC is​

Answer 1

Answer:

16.5×2 =33 cm

Step-by-step explanation:

use the property that tangents Drawn from a single point are equal in length

using that you will get that the perimeter of triangle is equal to twice the sum of BP,CQ AND ,AR.

Answer 2

(refer the attached pic for diagram)

given: A circle inscribed in ∆ABC touches sides

AB, BC, AC at P, Q, R respectively such that

BP = 6.5 cm, CQ = 4.5 cm amd AR = 5.5 cm

to find: perimeter of ∆ABC

solⁿ:

For the circle,

AP and AR is a pair of tangents from A to P and R

BP and BQ is a pair of tangents from B to P and Q

CQ and CR is a pair of tangents from C to Q and R

we know that, lengths of tangents from an external point, to a circle are equal.

Hence,

AP = AR = 5.5 cm ... (i) (given AR = 5.5 cm)

BQ = BP = 6.5 cm ... (ii) (given BP = 6.5 cm)

CR = CQ = 4.5 cm ... (iii) (given CQ = 4.5 cm)

now,

perimeter of ∆ABC

= AB + BC + AC

= (AP + BP) + (BQ + CQ) + (AR + CR)

= 5.5 + 6.5 + 6.5 + 4.5 + 5.5 + 4.5 ... (from i, ii and ii)

= 33 cm

Hence, perimeter of ∆ABC is 33 cm.